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Revision as of 21:28, 18 August 2005

Anonymous Coward: Hey, it sure seems to me that aleph-null fits the description of an inaccessible cardinal. Am I missing something?

No. ${\displaystyle \aleph _{0}}$ would satisfy the requirements for an inaccessible cardinal; that's why the definition specifically talks only about kardinals κ > ${\displaystyle \aleph _{0}}$. -- Schnee 05:59, 30 Sep 2004 (UTC)

This is a usage question. A few authors count ℵ₀ as an inaccessible, and even as a measurable. I think Drake does; don't have a copy here to check. --Trovatore 01:55, 14 July 2005 (UTC)

Rename or merge

By default, inaccessible means strongly inaccessible. The current page should be moved to "weakly inaccessible", and "strongly inaccessible" should be put here. Or better, the pages should be merged. --Trovatore 01:40, 14 July 2005 (UTC)

Merge complete. Result could still use some polishing--the two paragraphs on consistency strength could be usefully turned into one. --Trovatore 02:58, 14 July 2005 (UTC)

Consistency with ZFC

In fact, it cannot even be proven that the existence of strongly inaccessible cardinals is consistent with ZFC (as the existence of a model of ZFC + "there exists a strongly inaccessible cardinal" can be used to prove the consistency of ZFC)

I find this confusing. The existence of a model of ZFC + anything can be used to prove the consistency of ZFC, since such a model is also a model of ZFC. Josh Cherry 14:13, 31 July 2005 (UTC)

Good point. See fix. --Trovatore 15:03, 31 July 2005 (UTC)

OK, thanks. I'm not saying this is wrong, but I still don't get the argument. Proving that the existence of inaccessible cardinals is consistent with ZFC would, it seems from this, only show that the consistency of ZFC is consistent with ZFC, i.e., that if ZFC is consistent then ZFC + consis(ZFC) is consistent. Can't we prove that? Josh Cherry 15:14, 31 July 2005 (UTC)

No. If you could, then ZFC would prove that ZFC is consistent, contradicting Second Incompleteness. --Trovatore 15:31, 31 July 2005 (UTC)

Um, so I was slightly off here. If you could prove that Con(ZFC)-->Con(ZFC+Con(ZFC)), then ZFC+Con(ZFC) would prove its own consistency (and therefore be inconsistent). That's only minutely better or more plausible than ZFC itself being inconsistent. --Trovatore 16:01, 31 July 2005 (UTC)

Other descriptions

Is the first weakly inaccessible cardinal the same as ${\displaystyle \aleph _{\aleph _{\ddots }}}$? I've seen references to this, but it seems singular. Jim Apple 11:18, 18 August 2005 (UTC)

Well, depends on what you mean by that notation. Every weakly inaccessible cardinal is a fixed point of the ℵ function. I.e. if κ is weakly inaccessible then κ = ℵ_κ. But it's not the first fixed point, which is the limit of the ω-sequence

${\displaystyle \aleph _{0},\aleph _{\aleph _{0}},\aleph _{\aleph _{\aleph _{0}}},\ldots }$

You're quite right that this limit has cofinality ω --Trovatore 21:28, 18 August 2005 (UTC)